# The open box problem

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Introduction

The open box problem

Introduction

The aim of my algebraic investigation into the open box problem is to determine the size of square cut which makes the volume of the box as large as possible for any given rectangular sheet of card. The problem itself is simple, an open box is made from a sheet of card, identical squares are then cut off each of the four corners, the sheet is then folded to make box. It is my aim to find out the maximum square cut which gives me the maximum volume box.

Strategy

- Try to find the size of cut-out that will give me the maximum volume of a piece of card 6cm x 6cm, progressing onto algebra.
- Look for limitations in the results.
- Devise a plan for overcoming these limitations.
- Extend the work to find a general formula that will help me to work with all squares and rectangles.

Technique

In this situation there is no need for us to go even close to a pair of scissors or piece of card, for this investigation I am to use Excel. Excel is a computer programme in which I can input information; it will then calculate this information and give me results for what different size cut outs for different sized card.

The set-up of this is quite simple;

Size of card is out into cell c3 in this case

Volume is calculated

Middle

0.83

3.34

3.34

0.83

9.259148

0.84

3.32

3.32

0.84

9.258816

0.85

3.3

3.3

0.85

9.2565

Size of cut-out | Width | Depth | Height | Volume |

0.825 | 3.35 | 3.35 | 0.825 | 9.258563 |

0.826 | 3.348 | 3.348 | 0.826 | 9.25872 |

0.827 | 3.346 | 3.346 | 0.827 | 9.258857 |

0.828 | 3.344 | 3.344 | 0.828 | 9.258974 |

0.829 | 3.342 | 3.342 | 0.829 | 9.259071 |

0.830 | 3.34 | 3.34 | 0.83 | 9.259148 |

0.831 | 3.338 | 3.338 | 0.831 | 9.259205 |

0.832 | 3.336 | 3.336 | 0.832 | 9.259241 |

0.833 | 3.334 | 3.334 | 0.833 | 9.259258 |

0.834 | 3.332 | 3.332 | 0.834 | 9.259255 |

0.835 | 3.33 | 3.33 | 0.835 | 9.259232 |

With a 5cm x 5cm piece of card we’re able to see that a cut-out of 0.833 cm gives us the largest cut-out. A would not of necessarily been able to come to this conclusion if I don’t go ahead with a point that was brought forward in my plan and go to 3dp. If we now use the principle of my previous algebraic formula we can prove what we have seen.

V= [(5 – 2C) 2]x C

V= [(5 – 2x0.833) 2]x 0.833

V= 9.259258

It works, with graphing values into the formula we can see that we’ve got it right.

4cm x 4cm

Max Box | ||||

Size of card | 4 | |||

Size of cut-out | Width | Depth | Height | Volume |

0 | 4 | 4 | 0 | 0 |

1 | 2 | 2 | 1 | 4 |

2 | 0 | 0 | 2 | 0 |

3 | -2 | -2 | 3 | 12 |

Size of cut-out | Width | Depth | Height | Volume |

0.5 | 3 | 3 | 0.5 | 4.5 |

0.6 | 2.8 | 2.8 | 0.6 | 4.704 |

0.7 | 2.6 | 2.6 | 0.7 | 4.732 |

0.8 | 2.4 | 2.4 | 0.8 | 4.608 |

0.9 | 2.2 | 2.2 | 0.9 | 4.356 |

1.0 | 2 | 2 | 1 | 4 |

1.1 | 1.8 | 1.8 | 1.1 | 3.564 |

1.2 | 1.6 | 1.6 | 1.2 | 3.072 |

1.3 | 1.4 | 1.4 | 1.3 | 2.548 |

1.4 | 1.2 | 1.2 | 1.4 | 2.016 |

1.5 | 1 | 1 | 1.5 | 1.5 |

Size of cut-out | Width | Depth | Height | Volume |

0.65 | 2.7 | 2.7 | 0.65 | 4.7385 |

0.66 | 2.68 | 2.68 | 0.66 | 4.740384 |

0.67 | 2.66 | 2.66 | 0.67 | 4.740652 |

0.68 | 2.64 | 2.64 | 0.68 | 4.739328 |

0.69 | 2.62 | 2.62 | 0.69 | 4.736436 |

0.70 | 2.6 | 2.6 | 0.7 | 4.732 |

0.71 | 2.58 | 2.58 | 0.71 | 4.726044 |

0.72 | 2.56 | 2.56 | 0.72 | 4.718592 |

0.73 | 2.54 | 2.54 | 0.73 | 4.709668 |

0.74 | 2.52 | 2.52 | 0.74 | 4.699296 |

0.75 | 2.5 | 2.5 | 0.75 | 4.6875 |

Size of cut-out | Width | Depth | Height | Volume |

0.665 | 2.67 | 2.67 | 0.665 | 4.740719 |

0.666 | 2.668 | 2.668 | 0.666 | 4.740737 |

0.667 | 2.666 | 2.666 | 0.667 | 4.74074 |

0.668 | 2.664 | 2.664 | 0.668 | 4.740727 |

0.669 | 2.662 | 2.662 | 0.669 | 4.740697 |

0.670 | 2.66 | 2.66 | 0.67 | 4.740652 |

0.671 | 2.658 | 2.658 | 0.671 | 4.740591 |

0.672 | 2.656 | 2.656 | 0.672 | 4.740514 |

0.673 | 2.654 | 2.654 | 0.673 | 4.740421 |

0.674 | 2.652 | 2.652 | 0.674 | 4.740312 |

0.675 | 2.65 | 2.65 | 0.675 | 4.740188 |

With a 4cm x cm piece of card it is a cut-out of 0.667cm that gives us the maximum volume of 4.74074 cm3

V= [(4 – 2C) 2]x C

V= [(4 – 2x0.667) 2]x 0.677

V= 4.811815412

Although the calculation (and my calculator) give different answers, this could be because of simple problems and not because it’s wrong. In a situation like this, it usually comes down to decimal rounding and interpretation. Somewhere along the line either excel or my calculator have taken the decimal number and have changed this, because of this you get a differ of results, but close enough to see where your going. i.e turn the number into 1dp and you get 4.7 (1dp) which was the maximum original value to 1dp given when we looked at cut-out at 1dp.

3cm x 3cm

Max Box | ||||

Size of card | 3 | |||

Size of cut-out | Width | Depth | Height | Volume |

0 | 3 | 3 | 0 | 0 |

1 | 1 | 1 | 1 | 1 |

2 | -1 | -1 | 2 | 2 |

3 | -3 | -3 | 3 | 27 |

Size of cut-out | Width | Depth | Height | Volume |

0.0 | 3 | 3 | 0 | 0 |

0.1 | 2.8 | 2.8 | 0.1 | 0.784 |

0.2 | 2.6 | 2.6 | 0.2 | 1.352 |

0.3 | 2.4 | 2.4 | 0.3 | 1.728 |

0.4 | 2.2 | 2.2 | 0.4 | 1.936 |

0.5 | 2 | 2 | 0.5 | 2 |

0.6 | 1.8 | 1.8 | 0.6 | 1.944 |

0.7 | 1.6 | 1.6 | 0.7 | 1.792 |

0.8 | 1.4 | 1.4 | 0.8 | 1.568 |

0.9 | 1.2 | 1.2 | 0.9 | 1.296 |

1.0 | 1 | 1 | 1 | 1 |

Conclusion

And now I’m going to look at when K= 1/2 which is the way in which a rectangle is formed in the ration 2:1.

And it works, we find the cut-out to be of average 0.11A.

Conclusion

In this investigation I have been able to find out what and justify what I was originally set out to do, to find the maximum cut-out of any rectangular piece of card to give me the maximum volume. First of all I looked at square pieces of card and their actual measurements, volume after cat-out etc… From this information I was then able to look and concentrate on the measurements and come up with a general formula for a square piece of card, [(6-2C x C)2] x C I then went onto look at calculus with squares and was able to find out such things as the gradient function which is the moment where the gradient of the line, which the function belongs, is 0. Then I looked at rectangles with the ratio 1:2, forming a general formula again,

V= (1-2c)(2-2c)c for rectangles which I then used in calculus to help me find the answers to the volume with a k of 0.5. This algebraic way of working helped me to find a way in which I can find the cut-out of any piece of rectangular card which will give me the maximum volume.

This student written piece of work is one of many that can be found in our GCSE Open Box Problem section.

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